7.24

(a)

The relationship between number of calories and amount of carbs linear, represented by the trend up and to the right with a lot of outliers.

(b)

Number of calories is the explanatory variable. Amount of carbs is the response variable.

(c)

We might want to predict the amount of carbohydrate in Starbucks menu since Starbucks only list the number of calories on the display items.

(d)

Linearity: The relationship between number of calories and amount of carbs linear.

Nearly normal residuals: Residuals in this case have a slightly skeweed distribution to the left.

Constant variability: Based on the residual plot, we do not have constant variability for residuals.

Independent observations: Observations are independent.

We didn’t satisfy constant variability condition.

7.26

(a)

hMean = 171.14
hSD = 9.41
R = .67
gMean= 107.2
gSD= 10.37
slope = (hSD/gSD)*R
slope
## [1] 0.6079749
## hMean = y + (slope)*gMean
y = hMean - (slope)*gMean
y
## [1] 105.9651

height = 105 + 0.60 * shoulder girth

(b)

For eash additional cm in shoulder girth, the model predicts an additional 0.6 cm in height with base 105 cm at 0 shoulder girth, however it does not make sence if shoulder girth is zero.

(c)

R^2
## [1] 0.4489

About 45% of the variability in height is accounted for by the model.

(d)

height_student = 105 + 0.60 * 100
height_student
## [1] 165

(e)

e = 160-165
e
## [1] -5

A negative residual means that the model overestimate the height

(f)

height_child = 105 + 0.60 * 56
height_child
## [1] 138.6

It’s not possible that a one year old has 138 cm height I don’t think it would be appropriate to use this linear model to predict the height of children.

7.30

(a)

heart_weight = -0.357 + 4.034*body_weight

(b)

Expected heart_weight -0.357 obviously not a meaningful value, but it serves to adjust body_weight.

(c)

For each additional kg in body weight, we expect heart_weight to be higher on average by 4.034. Heart weight is measyred in g.

(d)

Body weight explains 64.66% of the variability in heart weight.

(e)

sqrt(0.6466)
## [1] 0.8041144

7.40

(a)

## 3.9983 = 4.010 + (-0.083*slope1)
slope1 = (3.9983 - 4.010)/-0.083
slope1
## [1] 0.1409639

(b)

Since the slope is positive the relationship is positive.

(c)

Linearity: There is a weak linear relationship.

Nearly normal residuals: Residuals in this case have a skeweed distribution to the left. Distribution of residuals is nearly normal.

Constant variability: The scatterplot of the residuals does appear to have constant variability.

Independent observations: Observations are independent.

All conditions are satisfied.